JEE Main & Advanced Sample Paper JEE Main - Mock Test - 13

  • question_answer
    A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is

    A) \[64.7V/m\]           

    B)       \[57.8V/m\]

    C) \[56.72V/m\]         

    D)        \[54.77V/m\]

    Correct Answer: D

    Solution :

    Intensity of EM wave is given by
    \[I=\frac{P}{4\pi {{R}^{2}}}={{v}_{a.v}}.c=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}\times c\]
    \[\Rightarrow \,\,\,\,{{E}_{0}}=\sqrt{\frac{P}{2\pi {{R}^{2}}{{\varepsilon }_{0}}c}}\]
    \[=\sqrt{\frac{800}{2\times 3.14\times {{(4)}^{2}}\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}}}\]
    \[=54.77\frac{V}{m}\]


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